Problem: $\dfrac{dy}{dt}=3y$ and $y(0)=2$. What is $t$ when $y=4$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $t=\ln \dfrac53$ (Choice B) B $t=0$ (Choice C) C $t=\dfrac{\ln 2}3$ (Choice D) D $t=\dfrac{\ln 3}2$ (Choice E) E $t=\dfrac{\ln 4}3$
Answer: The differential equation is separable. What does it look like after we separate the variables? $\dfrac{dy}y=3\,dt$ Let's integrate both sides of the equation. $\int\dfrac{dy}y=\int3\,dt$ What do we get? $\ln |y|=3t+C$ What value of $C$ satisfies the initial condition $y(0)=2$ ? Let's substitute $t=0$ and $y=2$ into the equation and solve for $C$. $\begin{aligned} \ln |2|&=3\cdot0+C\\ \\ C&=\ln 2 \end{aligned}$ Now use this value of $C$ to find $t$ when $y=4$. $\begin{aligned} \ln |4|&=3t+\ln 2\\ \\ 3t&=\ln 4-\ln 2\\ \\ 3t&=\ln \dfrac42\\ \\ \\ t&=\dfrac{\ln 2}3 \end{aligned}$